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I am allowed to consult other resources/individuals as long as I mention your name/position on the problem sheet. I have no idea how to even begin. This is well beyond the scope of my course. My instructor wants his students to interact/discuss this with the mathematics community.

Here it is:

Let Ω be a convex region in R2 and let L be a line segment of length ι that connects points on the boundary of Ω. As we move one end of L around the boundary, the other end will also move about this boundary, and the midpoint of L will trace out a curve within Ω that bounds a (smaller) region Γ. Find an expression that relates the area of Γ to the area of Ω in terms of the length ι.

crimsix
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  • there are a couple of papers on this will try to dig up references ... the first of these may have been from around 1850's – Mirko Dec 01 '14 at 19:01

2 Answers2

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The relevant result is known as Holditch theorem . From Wikipedia:
Holditch's theorem states that if a chord of fixed length is allowed to rotate inside a convex closed curve, then the locus of a point on the chord a distance $p$ from one end and a distance $q$ from the other is a closed curve whose area is less than that of the original curve by $\pi pq$.

His paper is
Rev. Hamnet Holditch, "Geometrical theorem"
The Quarterly Journal of Pure and Applied Mathematics 2, 1858, p. 38.
His one-page proof is beatiful.

I more modern (and more analytic) exposition is
Arne Broman, "A fresh look at a long-forgotten theorem",
Mathematics Magazine 54(3), May 1981, 99–108.

Mirko
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I suggest working it out for some simple $\Omega$.

If $\Omega$ is a circle of radius $R$, then $\Gamma$ is again a circle with radius $r<R$ that can be easily computed.

If $\Omega$ is a rectangle, then $\Gamma$ is the same rectangle with a quarter of circle taken out in each corner, see he picture below. Again computations can be made exactly.

enter image description here

In both cases $$ \text{area}(\Gamma)=\text{area}(\Omega)-\frac{\pi}{4}\,l^2. $$ That is, the area of $\Gamma$ is the area of $\Omega$ minus the area of a circle of radius $l/2$. Knowing the answer may show the way to the proof.

Julián Aguirre
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